The number of integral terms in the expansion | vedclass

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(B) The general term in the expansion of $(\sqrt{3} + \sqrt[8]{5})^{256}$ is given by $T_{r+1} = {}^{256}C_r (\sqrt{3})^{256-r} (\sqrt[8]{5})^r$.This

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$33$

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(B) The general term in the expansion of $(\sqrt{3} + \sqrt[8]{5})^{256}$ is given by $T_{r+1} = {}^{256}C_r (\sqrt{3})^{256-r} (\sqrt[8]{5})^r$.This simplifies to $T_{r+1} = {}^{256}C_r (3)^{\frac{256-r}{2}} (5)^{\frac{r}{8}}$.For the term to be an integer,both exponents $\frac{256-r}{2}$ and $\frac{r}{8}$ must be non-negative integers.Since $0 \leq r \leq 256$,for $\frac{r}{8}$ to be an integer,$r$ must be a multiple of $8$,i.e.,$r \in \{0, 8, 16, \dots, 256\}$.For these values of $r$,$\frac{256-r}{2} = 128 - \frac{r}{2}$ is also an integer because $r$ is a multiple of $8$ (and thus a multiple of $2$).The number of such values of $r$ is given by the arithmetic progression $0, 8, 16, \dots, 256$.Using the formula $a_n = a + (n-1)d$,we have $256 = 0 + (n-1)8$,which gives $n-1 = 32$,so $n = 33$.Thus,there are $33$ integral terms.

The number of integral terms in the expansion of $(\sqrt{3} + \sqrt[8]{5})^{256}$ is

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